Find the Electric Field at point P due to a finite rectangular sheet that contains a uniform
charge density σ.
z
b
r
P
r − r'
a
y
r'
x
r = x xˆ + y yˆ + z zˆ
r' = x ' xˆ + y ' yˆ
r − r' = ( x − x ') xˆ + ( y − y ') yˆ + z zˆ
For this problem, Cartesian coordinates would be the best choice in which to work the problem.
The electric field can be found using: E =
keσ dA
∫∫ r−r '
3 (r − r ') .
Since the sheet is in the xy-plane, the area element is dA = dx ' dy ' .
* Now we need expressions for & in terms of space (x, y, z) and body (x’, y’, z’) coordinates.
r − r ' = ( x − x ')xˆ + ( y − y ') yˆ + z zˆ
r − r ' = ( x − x ')2 + ( y − y ') 2 + z 2
(
r − r ' = ( x − x ') 2 + ( y − y ') 2 + z 2
3
)
3
2
∴
E = ∫∫
keσ dx ' dy '
(( x− x ')2 +( y − y ')2 + z2 ) 2
3
( x − x ') xˆ + ( y − y ') yˆ + z zˆ
This integral can be easier to deal with if it is broken down into component form:
E = E x xˆ + E y yˆ + E z zˆ
∴
Ex = keσ ∫∫
E y = keσ ∫∫
Ex = keσ ∫∫
( x − x ') dx ' dy '
(( x− x ')2 +( y − y ')2 + z2 ) 2
3
( y − y ') dx ' dy '
( ( x− x ')2 +( y − y ')2 + z2 ) 2
3
z dx ' dy '
(( x− x ') +( y − y ')2 + z2 ) 2
3
2
The limits on dx’ and dy’ are those that define the dimensions of the sheet:
a
a
≤ dx ' ≤
2
2
b
b
− ≤ dy ' ≤
2
2
−
∴
a
2
Ex = keσ ∫ a ( x − x ') dx ' ∫
−2
b
2
dy '
(
− b2 ( x − x ')2 + ( y − y ') 2 + z 2
a
2
E y = keσ ∫ a dx ' ∫
−2
a
2
b
2
−2
3
2
( y − y ') dy '
(
− b2 ( x − x ')2 + ( y − y ') 2 + z 2
Ez = keσ z ∫ a dx ' ∫
)
b
2
)
3
2
dy '
(
− b2 ( x − x ')2 + ( y − y ')2 + z 2
)
3
Note: z can be taken in front of both integrals
2
since it does not depend on x’ or y’.
* The dx’ integral can NOT be performed until the dy’ integral is evaluated since there is x’
dependence in the dy’ integral.
Comments:
* The analytical or closed form solution is extremely long and nasty for points off the z-axis
and will not be shown here.
* If P is at some point on the z-axis only, the Ex and Ey components vanish (due to symmetry)
and only the Ez component is left to evaluate (the solution is still somewhat nasty).